Question

In how many ways can 20 persons sit around a circle such that there is exactly one person between A and B?

• A. $18!\times 2!$
• B. $19!\times 2!$
• C. $18!\times 3!$
• D. $18!\times 4!$

Answer 1

The correct option is A $18!\times 2!$

We know that $n$ objects can be arranged around a circle in $(n-1)!$
If arranging these ${}^{\prime }{n}^{\prime }$ objects clockwise or counter clockwise means one and the same, then the number of arrangements will be half that number.
i.e., number of arrangements = $(n-1)!\times 2$
Let there be exactly one person between the two persons ($A$ and $B$) as stated in the question.
Then there will be $17$ others and this block of three people to be arranged around a circle.
The number of ways of arranging $18$ objects around a circle is in $17!$ ways.
Now $A$ and $B$ can be arranged on either side of the person sitting in between in $2!$ ways.
The person who sits between $A$ and $B$, can be any of the $18$ in the group and can be selected in $18$ ways.
Therefore, the total number of ways $18\times 17!\times 2!=18!\times 2!$